Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to proof that closed path $l$ (see picture) on 2-dimensional torus don't homotopic to trivial path, using definition torus as CW-complex? (closed path on surface $S$ is a map $f:[0, 1]\to S$, such that $f(0)=f(1)=x_0$) enter image description here Thanks.

share|cite|improve this question
It is difficult to give an answer that will be useful to you, without knowing your mathematical background. It would also be helpful to know where the question comes from. Is this for a class? Self-study? Do you have someone to talk to? What approaches have you tried? – Sam Nead Sep 15 '11 at 21:35
up vote 5 down vote accepted

Four proof sketches. $\newcommand{\ZZ}{\mathbb Z} \newcommand{\RR}{\mathbb R}$

  1. Note that the vertical loop (that you drew) and the horizontal loop have a single intersection point. Since the parity of the number of transverse intersection points is invariant under homotopy neither the vertical nor horizontal are homotopically trivial.

  2. The fundamental group of the torus is $\ZZ \times \ZZ$. The vertical and horizontal loops give the generators and so neither is trivial.

  3. Since the torus is the quotient $\RR^2/\ZZ^2$ we can take the preimage of the vertical loop in $\RR^2$ and get a collection of vertical lines. By homotopy lifting, a trivial loop in the torus lifts to an infinite collection of trivial loops in $\RR^2$.

  4. A loop in a surface is homotopically trivial if and only if it bounds a disk. The vertical loop is non-separating, so in particular does not bound a disk.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.